I'm looking for more information on the matrix that the Wronskian acts on to provide information about a set of solutions to an ODE.

So far I understand that:

  • The Wronskian comes from Cramer's rule (at least that is the explanation I see most of the time).
  • If the Wronskian is not equal to zero, then the set of solutions is linearly independent.
  • If the Wronskian is equal to zero, that the set of solutions could still be linearly independent, but likely isn't.
  • If the determinant of a matrix is equal to nonzero, then the matrix is linearly independent, it can be inverted and all that jazz.

What I'm confused on:

  • Why did we choose the derivatives of the function in the matrix inside of the Wronskian?
    • I've seen some... lame proofs that the derivatives of differentiable functions should be "linearly independent" to each other. I can easily think of functions that are differentiable that don't follow that rule (like $e^x$).
    • (This might be a stupid question): If the columns of the matrix are not linearly independent then isn't the entire matrix linearly dependent? Is this where our problems of the Wronskian doesn't guarantee linear dependence arise from?
  • 1
    $\begingroup$ It's not correct terminology to say that a matrix is linearly independent (if what you actually mean is that its columns are). $\endgroup$ Sep 11 at 7:51
  • $\begingroup$ And what do you mean by “the matrix that the Wronskian acts on”? $\endgroup$ Sep 11 at 7:52
  • $\begingroup$ You make a fair comment. The Wronskian is defined as the determinant of a matrix where each column is a function and its derivatives. What I guess what I'm asking about is the matrix that is used to calculate the Wronskian. Why do we use a matrix where the columns are the solutions, and their derivatives in the Wronskian? $\endgroup$
    – cardoza2
    Sep 13 at 17:23
  • $\begingroup$ OK, then it's clear what you mean. $\endgroup$ Sep 13 at 17:27

Solving for the coefficients of the solution to (at least) a second order differential equation

$$ y(x)=C_1u(x) + C_2v(x) $$ given the initial conditions: $y(t_0)=y_0$ and $y'(t_0)=y'_0$ means solving for the solution to the matrix below $$ \begin{bmatrix} u(t_0)&v(t_0) \\u'(t_0)&v'(t_0) \end{bmatrix} \begin{bmatrix} C_1\\C_2 \end{bmatrix} = \begin{bmatrix} y_0\\y'_0 \end{bmatrix} $$

which does not have a solution if the Wronskian is zero or the left most matrix has linearly dependent columns or rows (a row or column is a multiple of another row or column respectively).

In the interval $(a,b)$, the complete solution to the higher order linear differential equation must have $u,v$ to be linearly independent on $(a,b)$, the Wronskian not to be zero at some point $t_0$ in $(a,b)$ or never zero at all. If one of these conditions are met, the solution is called a fundamental solution set on the ODE at $(a,b)$.

  • $\begingroup$ Thanks for your answer. So what I get from your answer is the reason we check the linear independence of "the left most matrix" is because we want both the solutions and its derivatives to be linearly independent. $\endgroup$
    – cardoza2
    Sep 13 at 17:31
  • $\begingroup$ I had a comment about the Wronskian equaling zero in your answer, but I realized by comment was false. $\endgroup$
    – cardoza2
    Sep 13 at 17:36
  • $\begingroup$ I realized that it is important to note that we've assumed $u$ and $v$ are solutions of the ODE. Sometimes textbooks will say you can use the Wronskian on any set of functions, but the Wronskian is defined for a set of solutions. Thus if the Wronskian equals zero, then the set of solutions is linearly dependent. $\endgroup$
    – cardoza2
    Sep 13 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.